# Calculate Work to Drain a Hemispherical Tank

• Punkyc7
In summary: The work done is \pi\rho g \int_{-2}^0 (4- z^2)dz. That integral is (4z- z^3/3)|_{-2}^0= (4(0)- 0^3/3)- (4(-2)- (-2)^3/3)= 8- 16/3= 24/3- 16/3= 8/3= 2 2/3. But the question asked for the work needed to pump the water out of the tank, not just part way. The work needed to pump the water out of the tank is the work done to lift the water from depth z= -
Punkyc7
a bowl shaped tank is in the shaoe of a hemisphere with a radius of 2 in. If the bowl is filled with water density rho to a depth of 1 in, find the work in pumping the water out fo the tank

W=FD

V=integral of surface area * height

F= rho *V
V is the integral of pi(sqrt(4-y^2))^2 from 0-1
D=2-x

so W=rho*V int 2-x from 0-1

i get 11pi rho g/3 and the answer is suppose to be 9 rho pi/4

im not sure where i am going wrong but i thing it has to do with my integration

Assume that your hemi-spherical bowl has its flat surface in the xy-plane and the center of that surface at (0, 0, 0). Then $x^2+ y^2+ z^2= 4$

Now, imagine a "layer" of water at depth z (z< 0), of thickness "dz". The boundary of that layer is the circle $x^2+ y^2= 4- z^2$ which has radius $\sqrt{4- z^2}$ and so area $\pi(4- z^2)$ and volume $\pi(4- z^2)dz$. The weight of that water is $\pi\rho g(4- z^2)dz$ and we must lift it a height -z (remember that z is negative) out of the bowl. The work done is $-\pi\rho g(4- z^2)dz$.

Integrate that from z= -2 to z= 0.

## 1. How is work calculated for draining a hemispherical tank?

The work needed to drain a hemispherical tank can be calculated by using the formula W = mgh, where W is the work, m is the mass of the liquid being drained, g is the acceleration due to gravity, and h is the height of the liquid in the tank.

## 2. What is the formula for calculating the mass of liquid in a hemispherical tank?

The formula for calculating the mass of liquid in a hemispherical tank is m = ρV, where m is the mass, ρ is the density of the liquid, and V is the volume of the liquid.

## 3. How does the height of the liquid in the tank affect the work needed to drain it?

The height of the liquid in the tank directly affects the work needed to drain it. The higher the liquid level, the greater the potential energy it possesses, and therefore, the more work is needed to overcome the force of gravity and drain the liquid.

## 4. Does the shape of the tank affect the calculation of work?

Yes, the shape of the tank does affect the calculation of work. The formula for calculating work (W = mgh) takes into account the height of the liquid, which is affected by the shape of the tank. In the case of a hemispherical tank, the height of the liquid is equal to the radius of the tank, which is half of the diameter.

## 5. Can the work needed to drain a hemispherical tank be reduced?

Yes, the work needed to drain a hemispherical tank can be reduced by reducing the height of the liquid in the tank. This can be achieved by either draining some of the liquid before starting the draining process, or by using a smaller tank with a smaller volume of liquid.

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